UNIT 1 ASSESSMENT PROJECT Skylar Larson. 0.00 seconds 1.50 seconds1.00 seconds 0.50 seconds 3.50 seconds3.00 seconds2.50 seconds2.00 seconds

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<p>Particle Motion Project</p> <p>Unit 1 assessment projectSkylar Larson</p> <p>0.00 seconds1.50 seconds1.00 seconds0.50 seconds3.50 seconds3.00 seconds2.50 seconds2.00 seconds</p> <p>6.01905x^2-5.4x+3.55950.0 &lt; x &lt; 0.50-12.496696x^2+43.852139x-21.19176580.50 &lt; x &lt; 2.875-4.3429x^2+36.62286x-68.632.875 &lt; x &lt; 3.50f(x)=Edisplacement equations</p> <p>Height vs. Time GraphVelocity equationsf(x)=-24.99339x+43.85213912.0381x-5.4 0.50 &lt; x &lt; 2.875-8.6858x+36.62286 2.875 &lt; x &lt; 3.500.0 &lt; x &lt; 0.50Velocity vs. Time Graph</p> <p>Average Velocity T=0sT=3.5s</p> <p>3.5ft</p> <p>6.2ftAverage Velocity= (6.2 - 3.5)/(3.5 - 0)</p> <p>= 0.77143 ft/sInstantaneous Velocity at t=2.0secondsf(2) = -6.135 ft/sf(x)= -24.99339x+43.852139 0.50 &lt; x &lt; 2.875 2.0 seconds (From preview page)Instantaneous velocity at T=3.5 secondsf(x)= 6.2226 ft/s f(x)= -8.6858x+36.622862.875 &lt; x &lt; 3.50(From preview page)3.5 seconds Did the ball ever travel at 5 m/s (16.404 ft/s)?12.0381x-5.4 = 16.4040.0 &lt; x &lt; 0.50x=1.81125</p> <p>-24.99339x+43.852139 = 16.404 0.50 &lt; x &lt; 2.875 x=1.09822-8.6858x+36.62286 =16.404 2.875 &lt; x &lt; 3.50 x=2.3278f(x)= The ball will reach 5m/s (16.404 ft/s) at 1.09822 seconds. The other two x values are not in the domain. Part 2</p> <p>Definition of Derivative: lim = (f(x+h) f(x-h))/(2h) h 0 </p> <p>ft/sft/sInstantaneous rate of change at 2.0625 seconds2.0625 seconds</p> <p>Part 3f(x)=-2x+4 -1 &lt; x &lt; 1-(x-1)^2+2 1 &lt; x &lt; 4-0.5|x-8|+6 4 &lt; x &lt; 12Not continuous at x=4 because lim f(x)=4</p> <p>And lim f(x)=7</p> <p>x 4+x 4-They are not equalPart 3 </p> <p>f(x)=-2x+4 -1 &lt; x &lt; 1-(x-1)^2+2 1 &lt; x &lt; 4-0.5|x-8|-5 4 &lt; x &lt; 12Change function so there is a limit: move the absolute value equation/line down 11 units.Lim f(x) = -7 x 4Lim (x)= -7 x 4+-The limit does not exist at x = 4 because the left and right limits dont equal each other.Part 4Has a limit approaching infinity, as x approaches infinity.F(x)= (3x^2+4x)/(2x+7)Horizontal Asymptotes: None </p> <p>Part 4Has a limit approaching 0, as x approaches infinity.</p> <p>F(x)= (7x^2+3)/(2x^4+x)Horizontal Asymptotes: y=0 Part 4Has a limit approaching a line which is not 0, as x approaches infinity. </p> <p>F(x)= (3x^2+4x)/(4x^2)Horizontal Asymptotes: y= 0.75 (Found using coefficients)Part 4Has a limit approaching two separate lines as x approaches positive or negative infinity. </p> <p>F(x)= (|2x|)/(3x)Horizontal Asymptotes: y=2/3 and y= -2/3 (Found using coefficients)</p>